For a finite dimensional space, this holds for any two norms. Just compare the unit balls.
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DeaneJun 20 '13 at 18:15

Yes, "unit balls" is probably as good a reference as any. Hopefully, I can find a nice reference to close this question with. This is probably covered in some introductory chapter of an approximation theory book ...
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mathtickJun 23 '13 at 14:48

2 Answers
2

As I discussed in this answer, this inequality holds precisely when the measure space is atomic, i.e. $$\inf\{\mu(A) : \mu(A) > 0\} > 0.$$

I don't think it has a name, per se. One way to prove it is just to observe that, as in Markov's inequality, we have for any $a$,
$$\int f^2 \ge a^2 \mu(\{f \ge a\}).$$
For any $a < \|f\|_\infty$, we have $\mu(\{|f| \ge a\}) > 0$, hence $\mu(\{|f| \ge a\}) \ge m$. So for such $a$, $\int f^2 \ge a^2 m$. Now letting $a \uparrow \|f\|_\infty$ we get the desired result with $C = m^{-1/2}$.

If you like, you could write something like "by Markov's inequality" but that might just confuse the issue. I would probably write something like "because the measure is atomic, $\|f\|_\infty \le C \|f\|_2$" and leave it at that; the fact is very familiar for $\ell^p$ spaces (where $\mu$ is counting measure, so $m=1$) and the reader should be able to see that it holds for other atomic spaces. You can give the quick proof if you really feel it's necessary, and assign it an equation number if you need to refer to it often. It's a basic enough fact that I don't think a reference (or "simple online depiction", whatever that may be) is necessary.

In the finite dimensional case, as in Deane's comment, all norms are equivalent, and so I would just write "by equivalence of norms, $\|f\|_\infty \le C \|f\|_2$".